im trying to find the limit of 1.001^x as x approaches negative infinity. i know the answer is 0 because 1.001 raised to an decreasing negative number gives a value less than zero. but i wanted to know how would you show the work to prove that equal zero rather than just writing that the answer is zero

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anonymous · Answer

By "value less than zero" I think you mean either "value between 1 and 0" or something along the lines of a "value getting arbitrarily small and approaching 0." If you're looking for a somewhat rigorous approach to establishing that the limit is indeed 0 as $x\to-\infty$, you can focus on the sequence $a_n=\dfrac{1}{1.001^n}$ and show that $a_n$ is both decreasing and bounded below. You can establish these facts via induction. To prove the claim that $a_n$ is decreasing, you want to show that $a_n>a_{n+1}$ for all $n\ge1$. The basis case, that $a_1\dfrac{1}{1.001^2}$ because $1.001<1.001^2$. Assume this relation holds for $n=k$, i.e. that $a_k>a_{k+1}$, and use this to show that the relation holds for $n=k+1$, i.e. that $a_{k+1}>a_{k+2}$. $$a_{k+2}=\frac{1}{1.001^{k+2}}=\frac{1}{1.001}\times\frac{1}{1.001^{k+1}}<\frac{1}{1.001}\times\frac{1}{1.001^k}=\frac{1}{1.001^{k+1}}=a_{k+1}$$ So, $a_k>a_{k+1}$ implies $a_{k+1}>a_{k+2}$, and so $a_n$ is a decreasing sequence. Boundedness is a direct result of the fact that $a^b>0$ for any real $a>0$, $b\neq0$, so $a_n=\dfrac{1}{1.001^n}>0$ for all $n$. Taken together, the fact that the sequence is bounded below by 0 AND is decreasing, you can conclude that the limit of the sequence is 0, which proves the limit of your function is also 0. (The sequence $a_n$ is essentially the discrete form of your function.)